B5-2. Resistivity

1. Resistivity

Resistance is a macroscopic property; it depends heavily on an object's exact physical dimensions. However, resistivity ($\rho$) is the fundamental, microscopic property of the material itself. It quantifies how strongly a specific material opposes charge flow, independent of its shape.

Resistivity Equation:

$$R = \dfrac{\rho L}{A} \quad \Rightarrow \quad \rho = \dfrac{RA}{L}$$
  • $R$: Macroscopic resistance in Ohms (Ω).
  • $\rho$: Intrinsic resistivity in Ohm-metres (Ω m).
  • $L$: Length of the conductor (m).
  • $A$: Cross-sectional area (m²).

Example 1: The Volume Conservation Trap

Problem: A cylindrical piece of copper wire has an initial resistance of $4.0$ Ω. It is heated and mechanically drawn through a die so that its length is stretched to exactly three times its original length ($3L$). Calculate the new resistance of the wire, assuming no mass is lost.

Solution:

  • Step 1: Apply conservation of volume.
    Because mass and density are constant, the Volume ($V = A \times L$) must remain perfectly constant. If the length increases by a factor of 3 ($L_{\text{new}} = 3L$), the cross-sectional area must decrease by a factor of 3 to compensate ($A_{\text{new}} = A/3$).
  • Step 2: Apply the resistivity formula.
    $$R_{\text{new}} = \rho \dfrac{L_{\text{new}}}{A_{\text{new}}} = \rho \dfrac{3L}{(A/3)}$$
  • Step 3: Simplify and solve.
    $$R_{\text{new}} = 9 \left( \rho \dfrac{L}{A} \right) = 9 \times R_{\text{old}}$$
    $$R_{\text{new}} = 9 \times 4.0 = \mathbf{36.0 \text{ Ω}}$$
  • Universal Rule: Mechanically stretching a wire to $n$ times its length increases its resistance by a factor of $n^2$.

Example 2: Determining Microscopic Resistivity

Problem: An electrical engineer tests a specialized alloy wire of length $3.5$ m and diameter $0.50$ mm. When a potential difference of $2.4$ V is applied, a current of $0.80$ A is recorded. Calculate the resistivity of the alloy.

Solution:

  • Step 1: Calculate Resistance ($R$).
    $R = \dfrac{V}{I} = \dfrac{2.4}{0.80} = 3.0$ Ω
  • Step 2: Calculate Cross-Sectional Area ($A$).
    Radius $r = 0.25$ mm $= 0.25 \times 10^{-3}$ m.
    $A = \pi r^2 = \pi (0.25 \times 10^{-3})^2 \approx 1.963 \times 10^{-7}$ m²
  • Step 3: Calculate Resistivity ($\rho$).
    $$\rho = \dfrac{RA}{L} = \dfrac{(3.0) \times (1.963 \times 10^{-7})}{3.5} \approx \mathbf{1.68 \times 10^{-7} \text{ Ω m}}$$